Question

Aluminum in 2g of an ore sample is determined by dissolving it and then precipitating with
Al(OH)3 and igniting to Al2O3, which is weighed. What % of aluminum was in the sample if the ignited
precipitate weighed 0.2385g (Al = 27, O = 16, H = 1)?

Answer 1

I haven't got the reactions to hand but seems you've ended up with:

\(\dfrac{0.2385}{(2 \times 27) + (3 \times 16)}\) moles of the precipitate \(Al_2 O_3\)

so you had that many moles of Al in the sample, and those moles have mass:

\(\dfrac{0.2385}{(2 \times 27) + (3 \times 16)} \times 27 \text{ g}\)

And so the %-age you seek is:

\(\dfrac{ \dfrac{0.2385}{(2 \times 27) + (3 \times 16)} \times 27 }{2} \text{ %}\)